5
\$\begingroup\$

In "Instantaneous Power Theory and Applications to Power Conditioning" By Akagi, it states the following with regards to PQ Power Theory:

enter image description here

Is that \$\frac{1}{\sqrt3}\$ coefficient actually correct? Or is my approach incorrect?

I tried to derive by taking plugging in expressions from the Clarke matrix into the \$q(\alpha, \beta)\$ to find \$q(a,b,c)\$.

enter image description here enter image description here

And I've run through the calculation a few times now and I get a coefficient of \$\frac{1}{\sqrt2}\$. Essentially, what happens is that both terms contain the last row of each matrix as a multiplicand thereby imparting a coefficient of the \$\frac{\sqrt{3}}{2}\$ to the entire term and that coefficient immediately interacts the \$\sqrt{\frac{2}{3}}\$ coefficient in front of the matrix so there's nowhere for a \$\frac{1}{\sqrt3}\$ to come from.


EDIT: Hmmmmm...it looks like later (*actually, earlier if you look at the equation labels) the book justifies that:

enter image description here

by doing: enter image description here enter image description here

where: enter image description here

Adding that coefficient would make the result of my work match up with the initial statement but that doesn't really explain why what I did was invalid.


Here is the section as it is written: enter image description here enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Send an email to the author perhaps, but before that: check online if there may be an errata for that book. \$\endgroup\$ Commented Mar 24, 2022 at 20:16
  • \$\begingroup\$ @Kubahasn'tforgottenMonica See edit at the bottom. It seems like an ass pull though that isn't quite consistent. \$\endgroup\$
    – DKNguyen
    Commented Mar 24, 2022 at 20:16
  • \$\begingroup\$ @DKNguyen I see you added some stuff. Need to back off and read more. 3.1 looks right. Do you have Watanabe's 2008 paper in front of you? \$\endgroup\$
    – jonk
    Commented Mar 24, 2022 at 20:17
  • \$\begingroup\$ @jonk No, not the paper. I also just noticed that the part I asked about comes after the parts mentioned in my edit. Maybe I'm misinterpreting what the symbols are and somehow plugging the matrix expressions isn't allowed. To be honest, I'm not sure why 3.14 even has a coefficient since it seems to me it should be the same thing just with different basis. But whatever justification is used for it being there in 3.14 looks like it propagates over to the work I did above to get the result and since I didn't have it there earlier it doesn't work out. \$\endgroup\$
    – DKNguyen
    Commented Mar 24, 2022 at 20:30

1 Answer 1

1
\$\begingroup\$

As far as I know, the Clarke matrix has two versions: one that preserves the amplitude, \$C_a\$, and one that preserves power, \$C_p\$:

$$\begin{align} \pmb{C_p}=\sqrt{\dfrac23} &\begin{bmatrix} \dfrac{1}{\sqrt2} & \dfrac{1}{\sqrt2} & \dfrac{1}{\sqrt2} \\ 1 & -\dfrac12 & -\dfrac12 \\ 0 & \dfrac{\sqrt3}{2} & -\dfrac{\sqrt3}{2} \end{bmatrix} \tag{1} \\ \pmb{C_a}=\dfrac23 &\begin{bmatrix} \dfrac12 & \dfrac12 & \dfrac12 \\ 1 & -\dfrac12 & -\dfrac12 \\ 0 & \dfrac{\sqrt3}{2} & -\dfrac{\sqrt3}{2} \end{bmatrix} \tag{2} \end{align}$$

Now, if you consider his example further, you get these two results:

$$\begin{align} \pmb{v}&= \begin{cases} v_a(t)&=\sqrt2\sin(\omega t) \\ v_b(t)&=\sqrt2\sin(\omega t-\dfrac23\pi) \\ v_c(t)&=\sqrt2\sin(\omega t+\dfrac23\pi) \end{cases} \\ \pmb{C_p}\times\pmb{v}&= \begin{bmatrix} 0 \\ \sqrt3\sin(\omega t) \\ -\sqrt3\cos(\omega t) \end{bmatrix} \tag{3} \\ \pmb{C_a}\times\pmb{v}&= \begin{bmatrix} 0 \\ \sqrt2\sin(\omega t) \\ -\sqrt2\cos(\omega t) \end{bmatrix} \tag{4} \end{align}$$

So you can see that (3) adds the \$\sqrt3\$ factor which satisfies the equation (3.10), while (4) maintains the \$\sqrt2\$ as amplitude invariant. Now I see that the Wikipedia page also has the explanation, while Akagi's book (IIRC) does mention but only that the matrix used is power invariant. Therefore it makes sense to keep it since this is about powers.


I'm not sure if it was enough, so I'll add another example. Let's say that the voltages and the currents are defined in a generic way:

$$\begin{align} \pmb{v}&=[v_a,\;v_b,\;v_c]^T \tag{5a} \\ \pmb{i}&=[i_a,\;i_b,\;i_c]^T \tag{5b} \end{align}$$

Applying the power invariant transform gives the \$\alpha\beta\$ components (ignore the zero component):

$$\begin{align} \pmb{v_{\alpha\beta}}=\pmb{C_p}\times\pmb{v}&= \begin{bmatrix} \dfrac{2v_a-v_b-v_c}{\sqrt6} \\ \dfrac{v_b-v_c}{\sqrt2} \end{bmatrix} \tag{6a} \\ \pmb{i_{\alpha\beta}}=\pmb{C_p}\times\pmb{i}&= \begin{bmatrix} \dfrac{2i_a-i_b-i_c}{\sqrt6} \\ \dfrac{i_b-i_c}{\sqrt2} \end{bmatrix} \tag{6b} \end{align}$$

Now the power is calculated as \$q=v_{\beta}i_{\alpha}-v_{\alpha}i_{\beta}\$ and if you look at the denominators you'll see that you have \$\sqrt6\$ and \$\sqrt2\$, coupled with a \$2\$ in the numerator (one from \$v_a\$ and one that will come to be from \$\pm v_b\pm v_c\$). If you'll do the math (I'm tired of typing...), you'll end up with \$2/(2\sqrt3)=1/\sqrt3\$, as in (3.35).


For the sake of completeness, I'll expand on the previous example. I'll use \$i\cdot v\$ instead of \$v\cdot i\$ to make the factoring more clear and closer to equation (3.35) in the OP:

$$\begin{align}\require{cancel} q&=i_{\alpha}v_{\beta}-i_{\beta}v_{\alpha} \\ {}&=\dfrac{2i_a-i_b-i_c}{\sqrt6}\dfrac{v_b-v_c}{\sqrt2}-\dfrac{i_b-i_c}{\sqrt2}\dfrac{2v_a-v_b-v_c}{\sqrt6} \\ {}&=\dfrac{1}{2\sqrt3}(2i_av_b-2i_av_c-i_bv_b+i_bv_c-i_cv_b+i_cv_c-2i_bv_a+i_bv_b+i_bv_c+2i_cv_a-i_cv_b-i_cv_c) \\ {}&=\dfrac{1}{2\sqrt3}(2i_av_b-2i_av_c-\cancel{i_bv_b}+\cancel{i_bv_b}-\underline{2i_bv_a+i_bv_c+i_bv_c}+\overline{2i_cv_a-i_cv_b-i_cv_b}+\xcancel{i_cv_c}-\xcancel{i_cv_c}) \\ {}&=\dfrac{1}{2\sqrt3}\big[2i_a(v_b-v_c)-2i_b(v_a-v_c)+2i_c(v_a-v_b)\big] \\ {}&=\dfrac{2}{2\sqrt3}\big[i_a(v_b-v_c)-i_b(v_a-v_c)+ic(v(a-v_b)\big] \\ \Rightarrow \\ q&=\dfrac{1}{\sqrt3}\big[i_a(v_b-v_c)+i_b(v_c-v_a)+i_c(v_a-v_b)\big] \tag{7} \end{align}$$

\$\endgroup\$
7
  • \$\begingroup\$ I thought I was using Cp though as per 3.1 and 3.3 in my post but it stills comes out as though a \$\sqrt{\frac{2}{3}}\$ is missing. \$\endgroup\$
    – DKNguyen
    Commented Mar 24, 2022 at 21:47
  • 2
    \$\begingroup\$ [] I know what it is. I only have that coefficient in each term of q once but the each term contains two expressions derived from the Clarke matrix so the coefficient needs to be appear twice per term. So there needs to be two of them. That's why it seems like one is missing, because it is. [Edited by a moderator to remove profanity.] \$\endgroup\$
    – DKNguyen
    Commented Mar 24, 2022 at 21:57
  • \$\begingroup\$ @DKNguyen one last remark: if you'd be using the amplitude invariant matrix you'll end up with \$2/(3\sqrt3)\$ (and different \$\alpha\beta\$ components). \$\endgroup\$ Commented Mar 24, 2022 at 22:03
  • \$\begingroup\$ When do you choose one or the other? In the book they choose power invariant for voltages and currents because they are concerned with power, but it seems a little weird to choose power invariant and then apply it to the voltage and current. \$\endgroup\$
    – DKNguyen
    Commented Mar 24, 2022 at 22:12
  • 1
    \$\begingroup\$ @DKNguyen I think that is the point: voltage, by itself, has no power, same with current, but when multiplied... \$\endgroup\$ Commented Mar 24, 2022 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.